查看“︁放缩法”︁的源代码

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'''放缩法'''是通过舍去或添加一些项来构造[[不等式]]的一种方法。{{refn|group=参|{{cite web|url=http://www.math15.com/gaozhong/12_172.html|title=用放缩法证明不等式|deadurl=yes|archiveurl=https://web.archive.org/web/20140726021904/http://www.math15.com/gaozhong/12_172.html|archivedate=2014-07-26}} }}

==例子==
#求证<math>\sqrt{\log_2 3}+\sqrt{\log_3 2}<\sqrt{2}+1</math>{{refn|group=注|log为[[对数]]函数}}
#:<math>\sqrt{\log_2 3}+\sqrt{\log_3 2}<\sqrt{\log_2 4}+\sqrt{\log_3 3}=\sqrt{2}+1</math>{{refn|group=参|name=ref2|{{cite journal|author=董卫平|year=2011|title=说说放缩法|journal=数学大世界(高中)|issue=2|url=http://www.cnki.com.cn/Article/CJFDTOTAL-SXDJ201102012.htm|access-date=2015-09-20|archive-date=2016-03-04|archive-url=https://web.archive.org/web/20160304060910/http://www.cnki.com.cn/Article/CJFDTOTAL-SXDJ201102012.htm|dead-url=no}} }}
#已知a,b,c,d为[[正数]]，求证<math>1<\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}<2</math>
#:<math>\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1</math>
#:<math>\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}<\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d}=2</math>{{refn|group=参|name=ref3|{{cite web|url=http://www.pep.com.cn/rjwk/gzsxsxkj/2011/sxkj6/sxkj6gk/201109/t20110920_1069733.htm|title=例谈不等式证明的十种常用方法|accessdate=2014-07-16|archive-date=2014-07-20|archive-url=https://web.archive.org/web/20140720130452/http://www.pep.com.cn/rjwk/gzsxsxkj/2011/sxkj6/sxkj6gk/201109/t20110920_1069733.htm|dead-url=no}} }}
#求证<math>\sum_{k=1}^n \frac{1}{k^2}<2-\frac{1}{n}</math>
#:<math>\sum_{k=1}^n \frac{1}{k^2}<1+\sum_{k=2}^n \frac{1}{k(k-1)}=1+\sum_{k=2}^n \frac{1}{k-1}-\frac{1}{k}=2-\frac{1}{n}</math>{{refn|group=注|这里用了[[裂项和]]的求和方法}} {{refn|group=参|name=ref2}}
#设<math>n \in N^+</math>，求证<math> \frac{n(n+1)}{2}< \sum_{k=1}^n \sqrt{k(k+1)} < \frac{(n+1)^2}{2} </math>
#:<math>k=\sqrt{k^2}<\sqrt{k(k+1)}<\sqrt{k^2+k+\frac{1}{4}}=k+\frac{1}{2}</math>
#:<math>\frac{n(n+1)}{2}=\sum_{k=1}^n k < \sum_{k=1}^n \sqrt{k(k+1)} < \sum_{k=1}^n (k+\frac{1}{2}) =\frac{n^2+2n}{2}<\frac{(n+1)^2}{2}</math>{{refn|group=注|这里用了[[等幂求和]]的求和方法}} {{refn|group=参|name=ref3}}
#设a,b,c为[[直角三角形]]的三边,c为[[斜边]]，求证：<math>a^n+b^n<c^n(n>2)</math>
#:<math>a^n+b^n=a^2 a^{n-2}+b^2 b^{n-2}<a^2 c^{n-2}+b^2 c^{n-2}=c^n(n>2)</math>{{refn|group=注|这里用了[[勾股定理]]}} {{refn|group=参|name=ref2}}

==备注==
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==参考资料==
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[[分类:不等式]]